Finite Difference Method (FDM) home/lehre/vl-mhs-1-e/folien/vorlesung/2a_fdm/cover_sheet.tex page 1 of 15. p.1/15
Table of contents 1. Problem 2. Governing Equation 3. Finite Difference-Approximation 4. Formulation for each node 5. Stiffness Matrix 6. Analytical solution home/lehre/vl-mhs-1-e/folien/vorlesung/2a_fdm/table_of_contents.tex page 2 of 15. p.2/15
1D-Example - Finite Difference Method H left 1. Geometry A = 1 m 2 = 1m ents Q right x 2. Permeability k f = 10 5 m/s 1 m 1 4m 2 3 4 5 3. Boundary Conditions Q right = 10 4 m 3 /s H left = 50 m x x x x home/lehre/vl-mhs-1-e/folien/vorlesung/2a_fdm/problem.tex page 3 of 15. p.3/15
Governing Equations (FDM 1D) Continuity equation (mass balance equation for an incompressible fluid): v q = 0 Momentum balance equation Darcy equation (with all necessary assumptions): v = k f h with h = p ρg + z home/lehre/vl-mhs-1-e/folien/vorlesung/2a_fdm/governing_equation.tex page 4 of 15. p.4/15
Governing Equations (FDM 1D) If we insert the Darcy equation into the continuity equation and then rewrite it in a one dimensional form, we get ( k f h) q = 0 and 1D x ( k f Assumptions: No sinks or sources inside the tube, incompressible flow, and (Re 10). h x ) q = 0. home/lehre/vl-mhs-1-e/folien/vorlesung/2a_fdm/governing_equation2.tex page 5 of 15. p.5/15
Difference quotient (FDM 1D) Goal: Direct transfer from the partial differential equation to the difference equation (algebraic equation). Finite Difference Approximation h PSfrag replacements ĥ i ĥ i+1 (1) ĥ i 1 (2) (3) = 1.0[m] i 1 i i+1 home/lehre/vl-mhs-1-e/folien/vorlesung/2a_fdm/fd_approx1.tex page 6 of 15 x. p.6/15
Difference quotient (FDM 1D) We can write the finite difference formulation for the first derivative in three different ways. These can be deducted from a Taylor series expansion. ( ) dh dx i+ ( ) dh dx i ) ( dh dx i = h i+1 h i = h i h i 1 = h i+1 h i 1 2 forward difference (1) backward difference (2) central difference (3) home/lehre/vl-mhs-1-e/folien/vorlesung/2a_fdm/fd_approx2.tex page 7 of 15. p.7/15
Difference quotient (FDM 1D) The approximation of the second derivative can be obtained by summing the forward and the backward Taylor series. Thus, we eliminate the first derivative or we proceed as shown below by combining the first order terms. d 2 h dx 2 = d dx d 2 h dx 2 = 1 ( ) dh dx ( hi+1 h i = 1 h i h i 1 [( ) dh dx i+ ) ( ) ] dh dx i = h i+1 2h i + h i 1 2 home/lehre/vl-mhs-1-e/folien/vorlesung/2a_fdm/fd_approx3.tex page 8 of 15. p.8/15
Equations for each node (FDM 1D) For each unknown quantity we need one equation: Node 1 is our known boundary condition H left H left = h 1 or k f h 2 h 1 = q 1. Node 2 is calculated according to the presented scheme k f h i+1 2h i + h i 1 2 q 2 = 0 k f h 1 2h 2 + h 3 2 q 2 = 0, with q 2 = 0 we get, k f 1 2 (1, 2, 1) (h 1, h 2, h 3 ) T = 0, where h 1 is the b.c. home/lehre/vl-mhs-1-e/folien/vorlesung/2a_fdm/bal_eqn_node1.tex page 9 of 15. p.9/15
Equations for each node (FDM 1D) Node 3 and node 4 are calculated accordingly: k f 1 2 (1, 2, 1) (h 2, h 3, h 4 ) T = 0, k f 1 2 (1, 2, 1) (h 3, h 4, h 5 ) T = 0. Node 5 is a boundary node with the prescribed flux Q right. This boundary condition will be used directly as the last equation k f h 5 h 4 = Q right A k f 1 (1, 1) (h 4, h 5 ) T = Q right A. home/lehre/vl-mhs-1-e/folien/vorlesung/2a_fdm/bal_eqn_node2.tex page 10 of 15. p.10/15
Stiffness matrix (FDM 1D) Assemble the equations for the piezometric heads in matrix form 1 1 0 0 0 1 2 1 0 0 0 1 2 1 0 0 0 1 2 1 0 0 0 1 1 h 1 h 2 h 3 h 4 h 5 = k f q 1 0 0 0 k f q right. Here we use q right as boundary condition. q indicates specific discharges [m/s], while Q is discharge [m 3 /s]. home/lehre/vl-mhs-1-e/folien/vorlesung/2a_fdm/stiff_matrix1.tex page 11 of 15. p.11/15
Stiffness matrix (FDM 1D) As h 1 is known, we can delete the corresponding row and column and transfer h 1 to the right hand side of the equation. 2 1 0 0 1 2 1 0 0 1 2 1 0 0 1 1 h 2 h 3 h 4 h 5 = h left 0 0 k f q right This matrix can then be solved using for example a Gausselimination. You can try this at home! home/lehre/vl-mhs-1-e/folien/vorlesung/2a_fdm/stiff_matrix2.tex page 12 of 15. p.12/15
Analytical solution I The groundwater equation can be solved analytically if a number of assumptions are valid. In the one-dimensional case the equation looks as follows x ( k f h x ) q = 0. We assume now, that the permeability is not of function of x. We can therefore write it in front of the differential operator k f 2 h x 2 q = 0. In our case, there are no sources and sinks in the domain (q = 0). This yields the Laplace equation. home/lehre/vl-mhs-1-e/folien/vorlesung/2a_fdm/analy_sol_1.tex page 13 of 15. p.13/15
Analytical solution II The Laplace equation has the following form and can be solved by integrating the equation twice 2 h x 2 = 0 h x = C 1 h = C 1 x + C 2. Here the necessity of the boundary condition becomes clear. They are mandatory to evaluate the two constants C 1 and C 2. With the left boundary condition we calculate C 2 h(x = 0) = 50 m = C 1 0 + C 2 C 2 = 50 m. home/lehre/vl-mhs-1-e/folien/vorlesung/2a_fdm/analy_sol_2.tex page 14 of 15. p.14/15
Analytical solution III With the right boundary condition we can solve for C 1. In this case we have to connect the flow with the gradient of the piezometric head. To do so, we make use of Darcy s equation which we solve for h x Q = k f A h x, h x = C 1 = Q k f A = 10 4 m 3 /s 10 5 m/s 1 m = 10. 2 If we put all the information together we yield the piezometric head distribution. h(x) = 10x + 50 m. home/lehre/vl-mhs-1-e/folien/vorlesung/2a_fdm/analy_sol_3.tex page 15 of 15. p.15/15